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Mathematics, 18.03.2020 04:14 fatheadd2007

Find all pairs of real numbers (a, b) such that (x-a)^2+(2x-b)^2=(x-3)^2+(2x)^2

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Answer from: fendyli1529

(3,0)

$(\frac{-9}{5},\frac{12}{5})$

Step-by-step explanation:

Let's expand both sides.

I'm going to use the following identity to expand the binomial squared expressions: (u+v)^2=u^2+2uv+v^2 or (u-v)^2=u^2-2uv+v^2 .

Left-hand side:

(x-a)^2+(2x-b)^2

(x^2-2ax+a^2)+((2x)^2-2b(2x)+b^2)

x^2-2ax+a^2+4x^2-4bx+b^2

Reorder so x^2 's are together and that x[tex]'s are together.[tex](x^2+4x^2)+(-2ax-4bx)+(a^2+b^2)

5x^2+(-2a-4b)x+(a^2+b^2)

Right-hand side:

(x-3)^2+(2x)^2

x^2-2(3)x+9+4x^2

x^2-6x+9+4x^2

Reorder so x^2 's are together and that x[tex]'s are together.[tex](x^2+4x^2)+(-6x)+9

5x^2-6x+9

Now let's compare both sides.

If we want both sides to appear exactly the same we need to choose values and such the following are true equations:

-2a-4b=-6

a^2+b^2=9

So if we solve the system we can find the values and such that the left=right.

Let's solve the first equation for in terms of .

Add on both sides:

-4b=-6+2a

Divide both sides by -4:

$b=\frac{-6+2a}{-4}$

Reduce (divide top and bottom by -2):

$b=\frac{3-a}{2}$

Now let's plug this into second equation:

a^2+b^2=9

$a^2+(\frac{3-a}{2})^2=9$

$a^2+\frac{9-6a+a^2}{4}=9$ (I used the identity (u-v)^2=u^2-2uv+v^2 )

Multiply both sides by 4 to clear the fractions from the problem:

4a^2+(9-6a+a^2)=36

Combine like terms on left hand side:

4a^2+a^2-6a+9=36

5a^2-6a+9=36

Subtract 36 on both sides:

5a^2-6a-27=0

Now let's try to factor.

We are going to try to find two numbers that multiply to be 5(-27) and add to be -6.

5(-27)=(5*3)(-9)=15(-9)=-15(9) while -15+9=-6.

So let's replace -6a with -15a+9a and factor by grouping.

5a^2-15a+9a-27=0

5a(a-3)+9(a-3)=0

(a-3)(5a+9)=0

This implies a-3=0 or 5a+9=0 .

Solving the first is easy. Just ad 3 on both sides to get: a=3 .

The second requires two steps. Subtract 9 and then divide by 5 on both sides.

5a=-9

$a=\frac{-9}{5}$ .

So let's go back to finding now that we know the values.

If a=3 and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-3}{2}=0$ .

So one ordered pair (a,b) that satisfies the equation is:

(3,0) .

If $a=\frac{-9}{5}$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-\frac{-9}{5}}{2}$ .

Let's multiply top and bottom by 5 to clear the mini-fraction.

$b=\frac{15-(-9)}{10}$

$b=\frac{24}{10}$

$b=\frac{12}{5}$

So one ordered pair (a,b) that satisfies the equation is:

$(\frac{-9}{5},\frac{12}{5})$ .

ansver

Answer from: Quest

ab will be equal to

ansver

Answer from: Quest

d, x=8

step-by-step explanation:

simplifying

4x + -5 = 2x + 11

reorder the terms:

-5 + 4x = 2x + 11

reorder the terms:

-5 + 4x = 11 + 2x

solving

-5 + 4x = 11 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-5 + 4x + -2x = 11 + 2x + -2x

combine like terms: 4x + -2x = 2x

-5 + 2x = 11 + 2x + -2x

combine like terms: 2x + -2x = 0

-5 + 2x = 11 + 0

-5 + 2x = 11

add '5' to each side of the equation.

-5 + 5 + 2x = 11 + 5

combine like terms: -5 + 5 = 0

0 + 2x = 11 + 5

2x = 11 + 5

combine like terms: 11 + 5 = 16

2x = 16

divide each side by '2'.

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